ICould a <1g acceleration vehicle escape Earth's gravity well

Could a vehicle with less than 1G of thrust manage to escape the Earth's gravity well?

I'm thinking something like Space X's Falcon 9. As it landed, with just a little more thrust, it could go back up again, I would guess. So, if it could apply that tiny bit more thrust - and, assuming an infinite fuel supply - could the Falcon eventually leave Earth's gravity well?

I'm guessing that the thrust involved would be less than 1G since the delta v is certainly less than 9.8m/s; say, a few inches per minute. I mean, it's going up - at that few inches per minute - and, after enough time has passed, I'm guessing it could, eventually, reach a point where the Earth's gravity is negligible and will have escaped Earth's gravity well. But, it will never have reached escape velocity.

This seems wrong to me, but my head isn't wrapping around it properly.

it could, eventually, reach a point where the Earth's gravity is negligible and will have escaped Earth's gravity well. But, it will never have reached escape velocity.

Escape velocity is a function of altitude. The farther up you start, the lower the escape velocity is from that point. For any velocity, no matter how small, there is an altitude high enough so that the velocity exceeds escape velocity there.

Plus, once you get a few thousand miles up, the inverse square law will have reduced the earth's gravity so the rocket will be accelerating at a significant fraction of a gee. You'll get fast enough to exceed the surface escape velocity soon enough.

Staff: Mentor

Probably a minor quibble, but the way you worded it, it doesn't leave the launch pad. The acceleration from the ground can be anything, but just getting to zero acceleration (instead of falling down) requires "1G of thrust".

Or, a physicist would say that acceleration is what an accelerometer reads, so sitting on the ground it reads 1.0 and after leaving the ground it reads something above 1.0.

Could a vehicle with less than 1G of thrust manage to escape the Earth's gravity well?

What do you mean by "1G of thrust "? With thrust smaller than weight you cannot lift off vertically. But theoretically you could still speed up on some horizontal / inclined rail and reach escape velocity. Air drag makes this difficult on Earth though.

Escape velocity is a function of altitude. The farther up you start, the lower the escape velocity is from that point. For any velocity, no matter how small, there is an altitude high enough so that the velocity exceeds escape velocity there.

Plus, once you get a few thousand miles up, the inverse square law will have reduced the earth's gravity so the rocket will be accelerating at a significant fraction of a gee. You'll get fast enough to exceed the surface escape velocity soon enough.

So, if I'm reading this right, a ship with any acceleration could - given time - leave Earth? Even if that acceleration is less than Earth's gravity?

Probably a minor quibble, but the way you worded it, it doesn't leave the launch pad. The acceleration from the ground can be anything, but just getting to zero acceleration (instead of falling down) requires "1G of thrust".

Or, a physicist would say that acceleration is what an accelerometer reads, so sitting on the ground it reads 1.0 and after leaving the ground it reads something above 1.0.

....of course, as you move away from Earth it may fall below 1.0.

See, this is one part that throws me.

So, if the Falcon 9 is landing and, just before touching ground, it - assuming that it has a variable thrust - pushes out more thrust to make it move one inch higher over, say, ten minutes, then this would be a thrust of more than one gravity, even though the delta v is only 1-in/min and delta t is 10 minutes?

I think I'm taking the wrong path on my thinking, but I'm not sure where.

My basic thought is that a vessel with a motor that puts out 1G of thrust will not be able to leave the surface of Earth due to Earth's 1G of gravity pulling it down. The vector of the ship's thrust upwards is negated by the Earth's gravity vector downwards. The motor will have to put out over 1G to overcome gravity and lift off.

Is this what I'm seeing in my thought process on the hypothetical Falcon 9? That the ship's motor is already putting out 1G acceleration and the tiny bit more to get that 1-in of lift over 10 minutes is putting the motor's thrust to a small fraction more than 1G?

I think this question is a little off base. It isn’t achieving escape velocity that requires 1 G of thrust, it’s getting off the ground in the first place. By itself escape velocity doesn’t require a particular force. It requires accumulating enough energy. You are welcome to integrate for a very long time, so a small force will eventually produce escape velocity. For example once you’ve reached orbit and the forces opposing you are essentially zero any force greater than whatever slight drag remains applied for a long enough time will eventually accelerate you out of orbit. Escape velocity isn’t the limiting factor on the required thrust.

The problem is getting off the ground. The question isn’t can you reach escape velocity with a thrust less than 1 G. The question is can you reach orbit with a thrust less than 1 G. You can’t get off the ground without having greater than 1 G of upward force. If the only way to make lift is thrust, then we are done. For a rocket on the moon where going sideways isn’t an option thrust must be greater than 1 G. Does that mean you need 1 G of thrust? Well we get around that by using the atmosphere. Airplanes with much less than 1 G of thrust take off all the time.

Once in the air the plane has to accelerate against drag while maintaing lift >1 G. Can that be done with less than 1G of thrust? Well planes have managed to get to 100,000 ft and Mach 3 with less than 1 G of thrust. Both the lift and the drag drop off as the air gets thinner. Unfortunately so does the thrust. However if you could maintain 0.9 G of thrust indefinitely (and the mass wasn’t changing all the time from burning fuel) is there a path to transition to orbital speed and altitude without the drag exceeding 0.9 G? I don’t know. Maybe? Anyhow in that case the limit is lift and drag and has nothing to do with escape velocity.

Going back to the moon, if we made a magnetic levitation track around the moon’s equator so that there is essentially no drag, then a vehicle with any small thrust will eventually accelerate to orbital velocity and lift off the track and will continue accelerating to escape velocity.

What do you mean by "1G of thrust "? With thrust smaller than weight you cannot lift off vertically. But theoretically you could still speed up on some horizontal / inclined rail and reach escape velocity. Air drag makes this difficult on Earth though.

Basically, it means a thrust that equals the acceleration of Earth's gravity; 9.8m/s/s. 2G is 19.6m/s/s, 3G is 29.4m/s/s, etc.

I had an image I made that seemed to visually explain the rail idea. It was of a 1G ship taking off from a 1G planet using a launch rail.

It uses vectors to show the different forces as I see them and, admittedly, discounts friction and atmospheric drag.

The brown circle is the hypothetical rail.
The ship's acceleration vector is in Green, the planet's gravitational vector is in Blue.
The projected path is in Red and the actual path is in Black.

Basically, the rail acts as another vector acting against gravity by limiting the effect of the planet's gravity vector. Once the speed is high enough, the ship's resultant lifts it off the rail and gravity is insufficient to bring the ship down.

Granted, I've taken liberty concerning the actual path of the ship; I've ignored the curvature of the track and substituted the straight black vectors. But, if you shrink the time scale down, eventually the straight black vector is indistinguishable from the curved track, but the vector math would still apply.

Now, I'm not perfect. If anyone can show me any mistakes I've made here, I'll be happy to hear them - well, *read* them. The image seems correct, but my gut is still telling me a 1G ship is trapped on a 1G planet.

Basically, the rail acts as another vector acting against gravity by limiting the effect of the planet's gravity vector. Once the speed is high enough, the ship's resultant lifts it off the rail and gravity is insufficient to bring the ship down.

So you accelerate horizontally along the rails up to surface orbital velocity while still supported by the rails. Sure, that works. [If you can ignore that pesky air resistance]

Orbital velocity is the point where the curvature of the ship's path under the influence of gravity is equal to the curvature of the rails beneath it. Having achieved orbital velocity, continued thrust will eventually take you past escape velocity for the then-current altitude.

Staff: Mentor

Thrust is a force, not an acceleration. What you mean is thrust equal to weight.

While that's of course true, I think in this case, the shorthand/proxy is more useful. I doubt if during takeoff and landing real rockets even bother to calculate thrust, and instead control the fuel flow to maintain acceleration directly. Since both the mass of the rocket and thrust for a given fuel flow are constantly changing, it would be simpler to ignore both and control to the accelerometer.

And the physics error can easily be corrected with a minor grammatical tweak:

Instead of "1g of thrust" or "thrust of 1g", you just have to say "thrust for 1g". In the same way, helm commands in the Navy are given as "make turns [rpm] for 25 kts". The grid on the throttle reads in both knots and rpm, but knots are the more useful to know.

Staff: Mentor

So, if the Falcon 9 is landing and, just before touching ground, it - assuming that it has a variable thrust - pushes out more thrust to make it move one inch higher over, say, ten minutes, then this would be a thrust of more than one gravity, even though the delta v is only 1-in/min and delta t is 10 minutes?

Correct.

I think I'm taking the wrong path on my thinking, but I'm not sure where.

Honestly, you don't seem confused to me, so this is an odd place for the conversation to be....

My basic thought is that a vessel with a motor that puts out 1G of thrust will not be able to leave the surface of Earth due to Earth's 1G of gravity pulling it down. The vector of the ship's thrust upwards is negated by the Earth's gravity vector downwards. The motor will have to put out over 1G to overcome gravity and lift off.

Still correct!

Is this what I'm seeing in my thought process on the hypothetical Falcon 9? That the ship's motor is already putting out 1G acceleration and the tiny bit more to get that 1-in of lift over 10 minutes is putting the motor's thrust to a small fraction more than 1G?

If you are running the aircraft on a rail, the downward gravitational force from the earth is offset by the upward force from the rail. Your analysis is correct that if you accelerate at 1g or even much less than 1g for long enough, you can build up enough speed to have what is known as the "escape velocity". Even in the case without a rail, where you have flaps on the wings, with the upward force from the air pressure that can offset the force from the earth's gravity, what matters is the speed in the forward direction.## \\ ## (Edit: Additional item=in principle it requires no expenditure of energy or any work required to maintain a constant speed at a given altitude. In some cases, especially with atmospheric drag, engineering this is another matter. Outside of the atmosphere, one way to achieve this is to be in a circular orbit. In this sense. the gravitational force of ## F=mg ## does not need to be overcome. As previously mentioned, if the object is made to accelerate in its direction of motion, that acceleration increases its speed, and the escape velocity is the determining factor here ). ## \\ ## The acceleration of "1g" is not the hurdle here. The hurdle is the gravitational binding energy=see below. A stronger gravity=higher ## g ## goes hand-in-hand with a higher binding energy, but the actual hurdle is the binding energy=see below. ## \\ ## And the "escape velocity" comes about because the gravitational binding energy (at the earth's surface) ## U=-\frac{GM_e m}{R_e} ## is equal (and opposite) to the kinetic energy ## K.E.=\frac{1}{2} mv^2 ## when ## v ## is the escape velocity. The object has enough energy when ## v>v_{escape} ## that its trajectory ultimately results in being basically an infinite distance away where there is no gravitational pull, and if you compute the final velocity, you will find it still has some speed in the forward direction away from the earth, because the total energy is conserved. (## U=0 ## at ## r=+\infty ##. Meanwhile ## K.E._1+U_1=K.E._2+U_2= ## total energy, and total energy is positive if the object has the necessary escape velocity).